Find the Foci ((x-2)^2)/49+((y-4)^2)/25=1

$\frac{{(x - 2)}^{2}}{49} + \frac{{(y - 4)}^{2}}{25} = 1$

Step 1

Simplify each term in the equation in order to set the right side equal to $1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be $1$ .

$\frac{{(x - 2)}^{2}}{49} + \frac{{(y - 4)}^{2}}{25} = 1$

Step 2

This is the form of an ellipse. Use this form to determine the values used to find the center along with the major and minor axis of the ellipse.

$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$

Step 3

Match the values in this ellipse to those of the standard form. The variable $a$ represents the radius of the major axis of the ellipse, $b$ represents the radius of the minor axis of the ellipse, $h$ represents the x-offset from the origin, and $k$ represents the y-offset from the origin.

$a = 7$

$b = 5$

$k = 4$

$h = 2$

Step 4

Find $c$ , the distance from the center to a focus.

Find the distance from the center to a focus of the ellipse by using the following formula.

$\sqrt{a^{2} - b^{2}}$

Substitute the values of $a$ and $b$ in the formula.

$\sqrt{{(7)}^{2} - {(5)}^{2}}$

Simplify.

Raise $7$ to the power of $2$ .

$\sqrt{49 - {(5)}^{2}}$

Raise $5$ to the power of $2$ .

$\sqrt{49 - 1 \cdot 25}$

Multiply $- 1$ by $25$ .

$\sqrt{49 - 25}$

Subtract $25$ from $49$ .

$\sqrt{24}$

Rewrite $24$ as $2^{2} \cdot 6$ .

Factor $4$ out of $24$ .

$\sqrt{4 (6)}$

Rewrite $4$ as $2^{2}$ .

$\sqrt{2^{2} \cdot 6}$

Pull terms out from under the radical.

$2 \sqrt{6}$

Step 5

Find the foci.

The first focus of an ellipse can be found by adding $c$ to $h$ .

$(h + c, k)$

Substitute the known values of $h$ , $c$ , and $k$ into the formula.

$(2 + 2 \sqrt{6}, 4)$

The second focus of an ellipse can be found by subtracting $c$ from $h$ .

$(h - c, k)$

Substitute the known values of $h$ , $c$ , and $k$ into the formula.

$(2 - (2 \sqrt{6}), 4)$

Simplify.

$(2 - 2 \sqrt{6}, 4)$

Ellipses have two foci.

${Focus}_{1}$ : $(2 + 2 \sqrt{6}, 4)$

${Focus}_{2}$ : $(2 - 2 \sqrt{6}, 4)$

${Focus}_{1}$ : $(2 + 2 \sqrt{6}, 4)$

${Focus}_{2}$ : $(2 - 2 \sqrt{6}, 4)$

Step 6

Find the Foci ((x-2)^2)/49+((y-4)^2)/25=1

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